# Why does the short rate in the Hull White model follow a normal distribution?

Consider Hull White model $dr(t)=[\theta(t)-\alpha(t)r(t)]dt+\sigma(t)dW(t)$ when we solve the SDE above we have $r(t)=e^{-\alpha t}r(0)+\frac{\theta}{\alpha}(1-e^{-\alpha t})+\sigma e^{-\alpha t}\int_{0}^{t}e^{\alpha u}dW(u)$ and when we take expectation and variance we have $r(t) \sim N(e^{-\alpha t}r(0)+\frac{\theta}{\alpha}(1-e^{-\alpha t}),\frac{\sigma^2}{2\alpha}(1-e^{-\alpha t}))$.

I know the calculate how find SDE and find expectation or variance but I don't understand why $r(t)$ has normal distribution.

thanks.

For simplicity, we assume that $$\alpha$$ is a positive constant. You need to show that, for any $$t>0$$, \begin{align*} M_t = \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed, where $$\{W_t, \, t \ge 0\}$$ is a standard Brownian motion with respect to the filtration $$\{\mathscr{F}_t,\, t \ge 0\}$$. Here, we employ the time-changed Brownian motion technique. For $$t\ge 0$$, let $$\mathscr{G}_t = \mathscr{F}_{\frac{1}{2}\ln(1+2t)}$$. Consider the process $$X=\{X_t, t \geq 0\}$$, where \begin{align*} X_t = \int_0^{\frac{1}{2}\ln(1+2t)} e^{\alpha u} dW_u. \end{align*} Then $$X$$ is a continuous martingale with respect to the filtration $$\{\mathscr{G}_t,\, t \ge 0\}$$. Moreover, \begin{align*} \langle X, X\rangle_t &= \langle M, M\rangle_{\frac{1}{2}\ln(1+2t)}\\ &=\int_0^{\frac{1}{2}\ln(1+2t)} e^{2u} du =t. \end{align*} By Levy's martingale characterization of Brownian motion, $$\{X_t, t \ge 0\}$$ is a Brownian motion. That is, for $$t >0$$, $$X_t$$ is normally distributed. Consequently, for any $$t >0$$, \begin{align*} M_t &= \int_0^t e^{\alpha u} dW_u\\ &=X_{\frac{1}{2}(e^{2t}-1 )} \end{align*} is normally distributed, and $$r_t$$ is also normally distributed.

• $X_t$ is the product of a function of $t$ and an Ito integral wich is a martingale. How you say that $X_t$ is a martingale too?
– ab94
Sep 5 '19 at 17:48
• Thanks @ab94. There are indeed some mistakes in the previous answer. See the revision above. Nov 26 '19 at 19:20

This is a special case of the question of why $$\int_0^T f(t) dW_t$$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$\sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point is that the sum of independent normally distributed variables is again normally distributed.

• The second key point is that if a sequence of normal variable converges in law then the limit is also normal.
– AFK
May 18 '15 at 21:17
• The sum, which is normal, converges to the ito integral in $L^2(P)$, and the key point is that the $L^2$ limit of a sequence of normal random variables is still normal. May 18 '15 at 22:36